Proving that a trigonometric sum is in $L^2$

Question

How can I use Parseval's identity to prove that $$f(x)=\sum_{k=1}^\infty \frac{\sin(kx)}{1+k}$$ is in $L^2(0,\pi)$?

Thank you!

Answer 1

From Parseval's Theorem, we have

$$\frac{2}{\pi}\int_0^{\pi}f(x)^2dx=\sum_{k=1}^{\infty}\left(\frac{1}{k+1}\right)^2$$

Since the series converges, then $f$ is in $\mathscr{L}^2$

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We can show this equality as follows

$$\begin{align} \int_0^{\pi}f^2(x) &=\int_0^{\pi}f(x)\sum_{k=1}^{\infty}\frac{\sin kx}{k+1}dx \tag 1\\\\ &=\sum_{k=1}^{\infty}\frac{1}{k+1}\int_0^{\pi}f(x)\,\sin(kx)\,dx \tag 2\\\\ &=\sum_{k=1}^{\infty}\frac{1}{k+1}\frac{\pi}{2}\frac{1}{k+1} \tag 3\\\\ &=\frac{\pi}{2}\sum_{k=1}^{\infty}\left(\frac{1}{k+1}\right)^2 \end{align}$$

NOTE: 1:

$(1)$ uses the Fourier Series representation of $f$.

NOTE: 2:

In going from $(2)$ to $(3)$, we used the fact that the Fourier coefficients are given by $\frac{2}{\pi}\int_0^{\pi}f(x) \sin (kx)dx$.